The intervals are equinumerous

[evinda]

Hello! (Wave)

Each two open intervals in $\mathbb{R}$ are equinumerous.

It suffices to show that for each $a,b \in \mathbb{R}$ with $a<b$ it holds that $(0,1) \sim (a,b)$
$$f: (0,1) \xrightarrow[\text{surjective}]{\text{1-1}} (a,b)$$
$$f(t)=(1-t)a+b, t \in (0,1)$$

So this means for example that $(2,9) \sim (-5,100)$, right?
What surjective and 1-1 $f: (2,9) \to (-5,100)$ could we pick in order to show that these two intervals have the same cardinality?

 

[Evgeny.Makarov]

$$f: (0,1) \xrightarrow[\text{surjective}]{\text{1-1}} (a,b)$$
$$f(t)=(1-t)a+b, t \in (0,1)$$

It should be $f(t)=(1-t)a+tb$.

What surjective and 1-1 $f: (2,9) \to (-5,100)$ could we pick in order to show that these two intervals have the same cardinality?

You have two bijections: $f_1:(0,1)\to (2,9)$ and $f_2:(0,1) \to (-5,100)$. Then $f_2\circ f_1^{-1}: (2,9) \to (-5,100)$ is also a bijection.

 

[evinda]

It should be $f(t)=(1-t)a+tb$.

Oh yes, right! (Nod)

You have two bijections: $f_1: (0,1)\to (2,9)$ and $f_2: (0,1) \to (-5,100)$. Then $f_2\circ f_1^{-1}: (2,9) \to (-5,100)$ is also a bijection.

I see... Could we prove like that that $f$ is injective? (Thinking)

Let $x_1, x_2 \in (0,1)$ with $x_1 \neq x_2$.

$$x_1 \neq x_2 \Rightarrow bx_1 \neq bx_2 $$

$$x_1 \neq x_2 \Rightarrow -x_1 \neq -x_2 \Rightarrow 1-x_1 \neq 1-x_2 \Rightarrow (1-x_1) a \neq (1-x_2) a$$

$$\Rightarrow (1-x_1)a+x_1 b \neq (1-x_2)a+x_2b \Rightarrow f(x_1) \neq f(x_2)$$In order to prove that $f$ is surjective, we want to prove that $\forall y \in (a,b), \exists x \in (0,1)$ such that $f(x)=y \Rightarrow (1-x)a+xb=y$.
But how could we do this? (Thinking)

 

[Evgeny.Makarov]

Could we prove like that that $f$ is injective? (Thinking)

Let $x_1, x_2 \in (0,1)$ with $x_1 \neq x_2$.

$$x_1 \neq x_2 \Rightarrow bx_1 \neq bx_2 $$

This does not follow if $b=0$.

$$x_1 \neq x_2 \Rightarrow -x_1 \neq -x_2 \Rightarrow 1-x_1 \neq 1-x_2 \Rightarrow (1-x_1) a \neq (1-x_2) a$$

Again, the last implication does not hold if $a=0$.

$$\Rightarrow (1-x_1)a+x_1 b \neq (1-x_2)a+x_2b \Rightarrow f(x_1) \neq f(x_2)$$

And this does no hold at all. If you know that $a_1\ne a_2$ and $b_1\ne b_2$, it does not follow that $a_1+b_1\ne a_2+b_2$.

In order to prove that $f$ is surjective, we want to prove that $\forall y \in (a,b), \exists x \in (0,1)$ such that $f(x)=y \Rightarrow (1-x)a+xb=y$.
But how could we do this?

Have you tried solving a linear equation? This could help in proving injection as well.

 

[evinda]

If $a=0 \wedge b=0$ then we wouldn't take the function $f(t)=(1-t)a+tb$ since it wouldn't be injective, right?
In this case we could take the function $f(x)=x$, right?Now if $b=0, a \neq 0$ then $f(t)=(1-t)a$.

We pick $x_1, x_2 \in (0,1) $ with $x_1 \neq x_2$.

$x_1 \neq x_2 \Rightarrow -x_1 \neq -x_2 \Rightarrow 1-x_1 \neq 1-x_2 \Rightarrow (1-x_1) a \neq (1-x_2)a \Rightarrow f(x_1) \neq f(x_2)$

Now if $a=0, b \neq 0$ then $f(t)=tb$

We pick $x_1, x_2 \in (0,1) $ with $x_1 \neq x_2$.$x_1 \neq x_2 \Rightarrow bx_1 \neq bx_2 \Rightarrow f(x_1) \neq f(x_2)$

Now we consider the case $a \neq 0, b \neq 0$.

We pick $x_1, x_2 \in (0,1) $ with $x_1 \neq x_2$.

Suppose that $f(x_1)=f(x_2) \Rightarrow (1-x_1)a+bx_1=(1-x_2)b+b x_2 \Rightarrow b(x_1-x_2)=a(x_1-x_2) \overset{x_1 \neq x_2}{\Rightarrow} a=b$

When $a=b$ then $f(t)=a$ and that would mean that $f$ is not surjective, but we will show that it is.$$(1-x)a+xb=y \Rightarrow a-xa+xb=y \Rightarrow (b-a)x+a=y$$

$$a<y<b \Rightarrow a<(b-a)x+a<b \Rightarrow 0<(b-a)x<b$$Can we use this inequality to show that $f$ is surjective? (Thinking)

 

[Evgeny.Makarov]

$(0,1)$ and $(a,b)$ are not equinumerous when $a=b$.

We pick $x_1, x_2 \in (0,1) $ with $x_1 \neq x_2$.

Suppose that $f(x_1)=f(x_2) \Rightarrow (1-x_1)a+bx_1=(1-x_2)b+b x_2 \Rightarrow b(x_1-x_2)=a(x_1-x_2) \overset{x_1 \neq x_2}{\Rightarrow} a=b$

This is a proof of injectivity when $a\ne b$ regardless of whether $a=0$ or $b=0$.

 

[evinda]

$(0,1)$ and $(a,b)$ are not equinumerous when $a=b$.

Oh yes, right! (Nod)

This is a proof of injectivity when $a\ne b$ regardless of whether $a=0$ or $b=0$.

It always holds that $a \neq b$, so doing it like that we come to a contradiction, right? (Thinking)

 

[Evgeny.Makarov]

Yes.

 

[evinda]

Nice! (Smile)

And could we prove like that that $f$ is surjective? (Thinking)We want to show that $\forall y \in (a,b) \exists x \in (0,1)$ such that $f(x)=y$.
$f(x)=y \Rightarrow (1-x)a+bx=y \Rightarrow a-ax+bx=y \Rightarrow (b-a)x=y-a \overset{b \neq a}{\Rightarrow} x=\frac{y-a}{b-a}$

$$a<y<b \Rightarrow 0<y-a<b-a \overset{b-a>0}{\Rightarrow} 0< \frac{y-a}{b-a}<1 \Rightarrow 0<x<1$$So we see that we have shown that what we wanted to, therefore $f$ is surjective.

 

[Evgeny.Makarov]

Yes, that's correct.

 

[evinda]

Yes, that's correct.

Great! Thank you very much! (Happy)